Respuesta :
Using the z-distribution, we have that:
a)
- The 95% confidence interval estimate of the population proportion of customers who use their smartphone to access digital coupons is (0.4535, 0.5259).
- The 95% confidence interval estimate of the population proportion of customers who use their smartphone to access look-up recipes is (0.4494, 0.5218).
- The 95% confidence interval estimate of the population proportion of customers who use their smartphone to access read reviews is (0.2863, 0.3539).
- The 95% confidence interval estimate of the population proportion of customers who use their smartphone to access locate items is (0.1811, 0.2403).
b) Around 50% of the customers use their cell phones to either access digital coupons or look up recipes, while less than 50% use them to read reviews or locate items.
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
- z is the critical value.
- n is the sample size.
95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
Also, a sample of 731, hence n = 731.
Item a:
358 use their smartphone to access digital coupons, hence:
[tex]\pi = \frac{358}{731} = 0.4897[/tex]
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4897 - 1.96\sqrt{\frac{0.4897(0.5103)}{731}} = 0.4535[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4897 + 1.96\sqrt{\frac{0.4897(0.5103)}{731}} = 0.5259[/tex]
The 95% confidence interval estimate of the population proportion of customers who use their smartphone to access digital coupons is (0.4535, 0.5259).
355 use their smartphone to access look up recipes, hence:
[tex]\pi = \frac{355}{731} = 0.4856[/tex]
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4856 - 1.96\sqrt{\frac{0.4856(0.5144)}{731}} = 0.4494[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4856 + 1.96\sqrt{\frac{0.4856(0.5144)}{731}} = 0.5218[/tex]
The 95% confidence interval estimate of the population proportion of customers who use their smartphone to access look-up recipes is (0.4494, 0.5218).
234 use their smartphone to access read reviews of products, hence:
[tex]\pi = \frac{234}{731} = 0.3201[/tex]
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3201 - 1.96\sqrt{\frac{0.3201(0.6799)}{731}} = 0.2863[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3201 + 1.96\sqrt{\frac{0.3201(0.6799)}{731}} = 0.3539[/tex]
The 95% confidence interval estimate of the population proportion of customers who use their smartphone to access read reviews is (0.2863, 0.3539).
154 use their smartphone to access locate items, hence:
[tex]\pi = \frac{154}{731} = 0.2107[/tex]
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2107 - 1.96\sqrt{\frac{0.2107(0.7893)}{731}} = 0.1811[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2107 + 1.96\sqrt{\frac{0.2107(0.7893)}{731}} = 0.2403[/tex]
The 95% confidence interval estimate of the population proportion of customers who use their smartphone to access locate items is (0.1811, 0.2403).
Item b:
Considering if 50% = 0.5 is or is not part of the confidence interval, we have that:
Around 50% of the customers use their cell phones to either access digital coupons or look up recipes, while less than 50% use them to read reviews or locate items.
More can be learned about the z-distribution at https://brainly.com/question/25890103
